Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void R(int &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = (v + (v << 2) << 1) + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
inline void R(long double &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = v * 10 + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
bool vis[200005];
long double n;
long double ans[100005];
long double h, d;
int main() {
  cin >> n >> h;
  d = 100000;
  long double s = d * h / 2;
  long double temp;
  temp = s / n;
  for (long double i = 1; i < n; ++i) {
    printf("%0.14lf\n",
           (double)sqrt((double)i * (double)n) / (double)n * (double)h);
  }
  return 0;
}